lim-x-0-2-x-cos-x-sin-x-

Question Number 120425 by bobhans last updated on 31/Oct/20

\lim_{x \to 0} \frac{2^{x} - \cos x}{\sin x} ?

Answered by john santu last updated on 31/Oct/20

\lim_{x \to 0} \frac{2^{x} - \cos x}{\sin x} = \lim_{x \to 0} \frac{(2^{x} - \cos x) x}{x \sin x}

= \lim_{x \to 0} (\frac{2^{x} - \cos x}{x}) . \frac{x}{\sin x}

= \lim_{x \to 0} \frac{2^{x} - \cos x}{x} . \lim_{x \to 0} \frac{x}{\sin x}

= \lim_{x \to 0} \frac{2^{x} \ln 2 + \sin x}{1} . 1 = \ln 2 .

Answered by Lordose last updated on 31/Oct/20

\lim_{x \to 0} \frac{2^{x} - cosx}{sinx} = \lim_{x \to 0} \frac{2^{x} \ln 2 - sinx}{1} = \ln 2

Answered by Dwaipayan Shikari last updated on 31/Oct/20

\lim_{x \to 0} \frac{2^{x} - c o s x}{s i n x} = \lim_{x \to 0} \frac{2^{x} - 1}{x} (c o s x \to 1 s i n x \to x)

= \lim_{x \to 0} \frac{(x l o g (2) + 1 - 1)}{x} = l o g (2) \lim_{x \to 0} a^{x} = x l o g (a) + 1

Answered by mathmax by abdo last updated on 31/Oct/20

2^{x} = e^{xln 2} \sim 1 + xln 2, cosx \sim 1 - \frac{x^{2}}{2}

sinx \sim x - \frac{x^{3}}{6} \Rightarrow \frac{2^{x} - cosx}{sinx} \sim \frac{1 + xln 2 - 1 + \frac{x^{2}}{2}}{x - \frac{x^{3}}{6}}

= \frac{\ln 2 + \frac{x}{2}}{1 - \frac{x^{2}}{6}} \Rightarrow \lim_{x \to 0} \frac{2^{x} - cosx}{sinx} = \ln (2)